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If \(a\) and \(b\) are positive integers, is \(a^2 + b^2\) divisible by 5 ?

\(2ab\) is divisible by 5 \(a - b\) is divisible by 5

I have a question regarding this question. I want to verify the property of multiples that is related to divisibility. So

a) Mutliple of N + Mutliple of N = Multiple of Nb) Mutliple of N + Non -Mutliple of N = Non-Multiple of Nc) Non-Mutliple of N +Non- Mutliple of N = Can be both (multiple or non-multiple)

Does this property work for all integers, are there any exceptions?

There are no exceptions.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

As for the question:If \(a\) and \(b\) are positive integers, is \(a^2+b^2\) divisible by 5 ?

(1) \(2ab\) is divisible by 5 --> if \(a=b=5\) then the answer is YES but if \(a=5\) and \(b=1\) then the answer is NO. Not sufficient.

(2) \(a-b\) is divisible by 5 --> if \(a=b=5\) then the answer is YES but if \(a=b=1\) then the answer is NO. Not sufficient.

(1)+(2) From (2) \(a-b\) is divisible by 5 so \((a-b)^2=(a^2+b^2)-2ab\) is also divisible by 5. Next, since from (1) \(2ab\) is divisible by 5 then \(a^2+b^2\) must also be divisible by 5 in order their sum to be divisible by 5. Sufficient.